Question

Here is the Question of the Day !!!!!!!!!

An urn contains m white balls and n black balls .

A ball is drawn at random and is put back into the urn along with k additional balls of same colour.

A ball is drawn at random.

Show that probability of drawing a white ball does not depends on k.
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Answer 1

This is the given question is of

LAW OF TOTAL PROBABILITY.

Formula for LAW OF TOTAL PROBABILITY :-)

$P(A) = P(E_1).P( A | E_1)+ \\ P(E_2).P( A |E_2) + P(E_3).P( A| E_3) + ... .$

Solution for the Question is Shown in Attachment .

《 Refer to Attachment 》

Answer 2

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An urn contains m white balls and n black balls .A ball is drawn at random and is put back into the urn along with k additional balls of same colour.A ball is drawn at random.Show that probability of drawing a white ball does not depends on k.

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Let Event,

E_1 = First ball drawn of white colour

Event, E_₂ First ball drawn of black colour and Event, E Second ball drawn of white colour

$\sf \: P(E_1)= \frac{m}{m + n} \: \: and \: \: P(E_2) = \frac{n}{m + n}$

$\sf \: Also, P(E/E_1) = \frac{m}{m + n + k} \: \: and \: \: P(E/E_2) = \frac{m}{m + n + k}$

• Using total probability theorem,

$\sf \: P(E)=P(E_1)P(E/E_1) + P( E₂)P(E/E₂)$

$\sf \: \frac{m}{m + n} \times \frac{m + k}{m + n + k} \times \frac{n}{m + n} \times \frac{m}{m + n + k} \\ \\ \sf = \frac{m(m + k) + nm}{(m + n + k)(m + n)} \\ \\ \sf = \frac{m(m + k + n)}{(m + n + k)(m + n)} \\ \\ \sf = \frac{m}{m + n}$

Hence, the probability of drawing a white ball dies not depend on k.